Average induced EMF and average induced current of a flipped loop?

In summary: When the loop is flipped, what is the flux through the loop then? What is the difference?In summary, a wire loop with resistance R and area A is placed in a uniform magnetic field B, with its normal along the direction of the field. When the loop is flipped over in a time Δt, the normal becomes opposite to the field. The average induced EMF can be calculated using the equation Eind = ΔIB/Δt, where IB represents the total change of flux between the two positions. To find the average induced current, we can use the equation Iind = ΔEind/ΔR. In this case, the flux through the loop changes from BA to -BA, resulting in an average
  • #1
Violagirl
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Homework Statement


A wire loop of resistance R and area A has its normal along the direction of a uniform magnetic field, B. The loop is then flipped over in a time Δt so that its normal is opposite to the field. a) Calculate the average induced EMF. b) Find the average induced current. c) If the field is out of the page, what is the direction of the induced current? d) What is the total charge transported through the circuit in the interval Δt?


Homework Equations


IB = ∫NBA sin θ

εind = dIB/ dt

θ = ωt


The Attempt at a Solution



I've drawn the situation out on paper. In the beginning, it seems that B and A are uniform to one another at angle of cos θ = 0 degrees, which equals 1.

After the loop is flipped, B and A are no longer uniform as A will be downwards with B in the same, original direction. They will be an angle of cos θ = 180 degrees, which equals -1.

I also know that since they are looking for the average induced EMF, I will need to use the equation:

Eind = ΔIB/Δt

So I will need to find the separate EMF values of each position of the loop.

I know then that for the average induced current, it will be similar in that I will need:

Iind = ΔEind/ΔR

From here though, I'm confused on if I'm starting it correctly. In the first situation, I thought that B and A were uniform to one another so I took:

Ib = ∫NBA sin ωt

Ib = -NBAω cos ωt

so then Ib = NBAω

But then taking the derivative to find Eind is where I'm having trouble. If I wait to find the derivative of -NBAω cos ωt, we get NBAω sin ωt, which give an answer of 0 since sin 0 degrees is equal to 0 and it doesn't seem right...
 

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  • #2
Violagirl said:
Eind = ΔIB/Δt

So I will need to find the separate EMF values of each position of the loop.
There are not "separate EMF values of each position". There is a flux at each position. The EMF arises from the rate of change of flux.
Since you only want the average EMF, You don't need to worry about individual positions at all. ΔIB in your equation represents the total change of flux between the two positions. (To avoid confusion with current, I'll use ##\Phi## instead.) Since you only care about the average EMF over time Δt, you can just take the difference in flux between the two positions, ##\Delta\Phi##, and divide by Δt.
What is the flux through the loop in the initial position?
 

Related to Average induced EMF and average induced current of a flipped loop?

1. What is average induced EMF?

Average induced EMF (Electromotive Force) is the average value of the electric potential difference induced in a loop of wire when it is rotated or moved in a magnetic field.

2. How is average induced EMF calculated?

The average induced EMF is calculated by multiplying the number of turns in the loop by the rate of change of magnetic flux through the loop.

3. What factors affect the average induced EMF?

The average induced EMF is affected by the strength of the magnetic field, the speed of movement of the loop, and the number of turns in the loop.

4. What is average induced current?

Average induced current is the average flow of electric charge in a loop of wire when it is moved or rotated in a magnetic field, due to the induced EMF.

5. How is average induced current related to average induced EMF?

The average induced current is directly proportional to the average induced EMF. As the induced EMF increases, so does the induced current.

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